Lecture 05: Probability and Statistical Inference

Bill Perry

Lecture 4: Review

Introduction to histograms or frequency distributions
Probability Distribution Functions (PDF)
Descriptive Statistics
- Center - mean, median, mode
- Spread - range, variance, standard deviation

Lecture 4: Review

Introduction to histograms or frequency distributions
Probability Distribution Functions (PDF)
Descriptive Statistics
- Center - mean, median, mode
- Spread - range, variance, standard deviation

Lecture 4: Review

Introduction to histograms or frequency distributions
Probability Distribution Functions (PDF)
Descriptive Statistics
- Center - mean, median, mode
- Spread - range, variance, standard deviation

library(tidyverse)
library(patchwork)
library(car)          # For diagnostic tests

grayling_df <- read_csv("data/gray_I3_I8.csv") 
i3_df  <- grayling_df %>% filter(lake=="I3")

# Calculate summary statistics
grayling_summary <- grayling_df %>% 
  group_by(lake) %>%
  summarize(
    mean_length = mean(length_mm, na.rm = TRUE),
    sd_length = sd(length_mm, na.rm = TRUE),
    se_length = sd_length/sqrt(sum(!is.na(length_mm))),
    count = sum(!is.na(length_mm)),
    .groups = "drop")
grayling_summary

# A tibble: 2 × 5
  lake  mean_length sd_length se_length count
  <chr>       <dbl>     <dbl>     <dbl> <int>
1 I3           266.      28.3      3.48    66
2 I8           363.      52.3      5.18   102

Lecture 5: Probability and Statistical Inference

The goals for today

Statistical inference fundamentals
Hypothesis testing principles
T Distributions
One sample T Tests
Two sample T Test

Lecture 5: Confidence intervals

In the more typical case DON’T know the population σ or standard deviation

estimate it from the samples
and when sample size is <~30)
can’t use the standard normal (z) distribution

Instead, we use Student’s t distribution

Lecture 5: Understanding t-distribution

When sample sizes are small, the t-distribution is more appropriate than the normal distribution.

Similar to normal distribution but with heavier tails
Shape depends on degrees of freedom (df = n-1)
With large df (>30), approaches the normal distribution
Used for:
- Small sample sizes
- When population standard deviation is unknown
- Calculating confidence intervals
- Conducting t-tests

Lecture 5: Understanding t-distribution

When sample sizes are small, the t-distribution is more appropriate than the normal distribution.

Similar to normal distribution (1.96 = 2.5% tails) but with heavier tails
Shape depends on degrees of freedom (df = n-1)
With large df (>30), approaches the normal distribution
Used for:
- Small sample sizes
- When population standard deviation is unknown
- Calculating confidence intervals
- Conducting t-tests

Practice Exercise 4: Using the t-distribution

Let’s compare confidence intervals using the normal approximation (z) versus the t-distribution for our fish data.

# Calculate CI using both z and t distributions for a smaller subset
small_sample <- grayling_df %>% 
  filter(lake == "I3") %>% 
  slice_sample(n = 10)

# Calculate statistics
sample_mean <- mean(small_sample$length_mm)
sample_sd <- sd(small_sample$length_mm)
sample_n <- nrow(small_sample)
sample_se <- sample_sd / sqrt(sample_n)

# Calculate confidence intervals
z_ci_lower <- sample_mean - 1.96 * sample_se
z_ci_upper <- sample_mean + 1.96 * sample_se

# For t-distribution, get critical value for 95% CI with df = n-1
t_crit <- qt(0.975, df = sample_n - 1)
t_ci_lower <- sample_mean - t_crit * sample_se
t_ci_upper <- sample_mean + t_crit * sample_se

# Display results
cat("Mean:", round(sample_mean, 1), "mm\n")

Mean: 273.3 mm

cat("Standard deviation:", round(sample_sd, 2), "mm\n")

Standard deviation: 38.54 mm

cat("Standard error:", round(sample_se, 2), "mm\n")

Standard error: 12.19 mm

cat("95% CI using z:", round(z_ci_lower, 1), "to", round(z_ci_upper, 1), "mm\n")

95% CI using z: 249.4 to 297.2 mm

cat("95% CI using t:", round(t_ci_lower, 1), "to", round(t_ci_upper, 1), "mm\n")

95% CI using t: 245.7 to 300.9 mm

cat("t critical value:", round(t_crit, 3), "vs z critical value: 1.96\n")

t critical value: 2.262 vs z critical value: 1.96

Student’s t-distribution

To calculate CI for sample from “unknown” population:

\(\text{CI} = \bar{y} \pm t \cdot \frac{s}{\sqrt{n}}\)

Where:

ȳ is sample mean
𝑛 is sample size
s is sample standard deviation
t t-value corresponding the probability of the CI
t in t-table for different degrees of freedom (n-1)

Lecture 5: Student’s t-distribution

Here is a t-table

Values of t that correspond to probabilities
Probabilities listed along top
Sample dfs are listed in the left-most column
Probabilities are given for one-tailed and two-tailed “questions”

Lecture 5: Student’s t-distribution

One-tailed questions: area of distribution left or (right) of a certain value

n=20 (df=19) - 90% of the observations found left
t= 1.328 (10% are outside)

Lecture 5: Student’s t-distribution

Two-tailed questions refer to area between certain values

n= 20 (df=19), 90% of the observations are between
t=-1.729 and t=1.729 (10% are outside)

Lecture 5: Student’s t-distribution

Let’s calculate CIs again:

Use two-sided test

\(\text{CI} = \bar{y} \pm t \cdot \frac{s}{\sqrt{n}}\)
95% CI Sample A: = 272.8 ± 2.262 * (37.81/(9^0.5)) = 1.650788
The 95% CI is between 244.3 and 301.3
“The 95% CI for the population mean from sample A is 272.8 ± 28.5”

Lecture 5: Student’s t-distribution

So:

Can assess confidence that population mean is within a certain range
Can use t distribution to ask questions like:
- “What is probability of getting sample with mean = ȳ from population with mean = µ?” (1 sample t-test)
- “What is the probability that two samples came from same population?” (2 sample t-test)

Lecture 5: Single Sample T-Test

We want to test if the mean fish length in I3 differs from 240mm.

Activity: Define hypotheses and identify assumptions

H₀: μ = 240 (The mean fish length in I3 is 240mm)

H₁: μ ≠ 240 (The mean fish length in I3 is not 55mm)

Assumptions for t-test:

Data is normally distributed
Observations are independent
No significant outliers

Assumptions in R - qqplots from car

# Filter for just windward side needles

# YOUR TASK: Test normality of windward pine needle lengths
# QQ Plot
qqPlot(i3_df$length_mm, 
       main = "QQ Plot for length of Grayling",
       ylab = "Sample Quantiles")

[1] 53 35

Statistical Test of Normality - Shapiro-Wilk test

# Shapiro-Wilk test
shapiro_test <- shapiro.test(i3_df$length_mm)
print(shapiro_test)


    Shapiro-Wilk normality test

data:  i3_df$length_mm
W = 0.91051, p-value = 0.0001623

# Check for outliers using boxplot
# YOUR CODE HERE
i3_df %>% ggplot(aes(lake, length_mm))+geom_boxplot()

Now for Practice

Practice Exercise 1: One-Sample t-Test

Let’s perform a one-sample t-test to determine if the mean fish length in I3 Lake differs from 240 mm:

# what is the mean
i3_mean <- mean(i3_df$length_mm, na.rm=TRUE)
cat("Mean:", round(i3_mean, 1), "mm\n")

Mean: 265.6 mm

# Perform a one-sample t-test
t_test_result <- t.test(i3_df$length_mm, mu = 240)

# View the test results
t_test_result


    One Sample t-test

data:  i3_df$length_mm
t = 7.3497, df = 65, p-value = 4.17e-10
alternative hypothesis: true mean is not equal to 240
95 percent confidence interval:
 258.6481 272.5640
sample estimates:
mean of x 
 265.6061

Interpret this test result by answering these questions:

What was the null hypothesis?
What was the alternative hypothesis?
What does the p-value tell us?
Should we reject or fail to reject the null hypothesis at α = 0.05?
What is the practical interpretation of this result for fish biologists?

Lecture 5: Intro to Hypothesis Testing

Hypothesis testing is a systematic way to evaluate research questions using data.

Key components:

Null hypothesis (Ho): Typically assumes “no effect” or “no difference”
Alternative hypothesis (Ha): The claim we’re trying to support
Statistical test: Method for evaluating evidence against H₀
P-value: Probability of observing our results (or more extreme) if H₀ is true
Significance level (α): Threshold for rejecting H₀, typically 0.05

Decision rule: Reject Ho if p-value < α == p < 0.05

Lecture 5: Intro to Hypothesis Testing

Hypothesis testing is a systematic way to evaluate research questions using data.

Key components:

Null hypothesis (H₀): Typically assumes “no effect” or “no difference”
Alternative hypothesis (Hₐ): The claim we’re trying to support
Statistical test: Method for evaluating evidence against H₀
P-value: Probability of observing our results (or more extreme) if H₀ is true
Significance level (α): Threshold for rejecting H₀, typically 0.05

Decision rule: Reject H₀ if p-value < α

Lecture 5: Interpreting and Reporting Results of a 1 sample T Test

Activity: Interpret the t-test results

What does the p-value tell us?
Should we reject or fail to reject the null hypothesis?

How to report this result in a scientific paper:

“A two-tailed, one-sample t-test at α=0.05 showed that the mean pine needle length on the windward side (… mm, SD = …) [was/was not] significantly different from the expected 55 mm, t(…) = …, p = …”

Lecture 5: Next steps - two sample T Tests

For example

what is probability that population X is the same as population Y?

How would you assess this question using what we learned?

This is what we will do with the pine needles…

Lecture 5: Next steps - two samples

For example

what is probability that population X is the same as population Y?

How would you assess this question using what we learned?

# Now create a boxplot to visualize the difference in fish lengths between these lakes:
pine_df <- read_csv("data/pine_needles.csv")

# Create a boxplot comparing the two lakes
pine_wind_plot <- pine_df %>%
  ggplot(aes(x = wind, y = length_mm, fill = wind)) +
  geom_boxplot() +
  labs(title = "Pine Needle Lengths by Wind Exposure",
       x = "Position",
       y = "Length (mm)",
       fill = "Wind Position") +
  scale_fill_manual(values = c("lee" = "forestgreen", "wind" = "skyblue"),
                   labels = c("lee" = "Leeward", "wind" = "Windward"))
pine_wind_plot

# Based on the t-test results and the boxplot
# 
# what can you conclude about the fish populations in these two lakes?

Practice Exercise 2: Formulating Hypotheses

For the following research questions about pine needles write the null and alternative hypotheses:

Are needles on the lee side longer than the needles on the windy side?

What are the hypotheses?

Ho =

Ha =

Lecture 5: Next steps - two samples

Now, let’s compare pine needle lengths between windward and leeward sides of trees.

Question: Is there a significant difference in needle length between the windward and leeward sides?

This requires a two-sample t-test.

Two-sample t-test compares means from two independent groups.

\(t = \frac{\bar{x}_1 - \bar{x}_2}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\)

where:

x̄₁ and x̄₂: These represent the sample means of the two groups you’re comparing
s²ₚ: This is the pooled variance, calculated as: s²ₚ = [(n₁ - 1)s₁² + (n₂ - 1)s₂²] / (n₁ + n₂ - 2), where s₁² and s₂² are the sample variances of the two groups.
n₁ and n₂: These are the sample sizes of the two groups.
√(1/n₁ + 1/n₂): This represents the pooled standard error.

\(t = \frac{SIGNAL}{NOISE}\)

Practice Exercise 3: Calculate summary statistics grouped by wind exposure

Before conducting the test, we need to understand the data for each group.

You need this and the graph to see what is goin on ….

group_summary <- pine_df %>%
  group_by(wind) %>%
  summarize(
    mean_length = mean(length_mm),
    sd_length = sd(length_mm),
    n = n(),
    se_length = sd_length / sqrt(n)
  )

print(group_summary)

# A tibble: 2 × 5
  wind  mean_length sd_length     n se_length
  <chr>       <dbl>     <dbl> <int>     <dbl>
1 lee          20.4      2.45    24     0.500
2 wind         14.9      1.91    24     0.390

# Create a boxplot comparing the two sides
pine_wind_plot

Practice Exercise 4: Effect size

We could also look at the difference in means… some cool code here

# Assuming your dataframe is called df
group_summary %>%
  summarize(difference = mean_length[wind == "wind"] - mean_length[wind == "lee"])

# A tibble: 1 × 1
  difference
       <dbl>
1       -5.5

Practice Exercise 5: Using GGPLOT to get summary stats

GGplot also has code to make the mean and standard error plots we are interested in along whit a lot of others

# Assuming your dataframe is called df
pine_mean_se_plot <- ggplot(pine_df, aes(x = wind, y = length_mm, color = wind)) +
  stat_summary(fun = mean, geom = "point") +
  stat_summary(fun.data = mean_se, geom = "errorbar", width = 0.2) +
  labs(title = "Mean Pine Needle Length by Wind Exposure",
       x = "Wind Exposure",
       y = "Mean Length (mm)") +
  coord_cartesian(ylim = c(0,25))+
  scale_color_manual(values = c("lee" = "forestgreen", "wind" = "skyblue"),
                   labels = c("lee" = "Leeward", "wind" = "Windward"))+
  theme_classic()
pine_mean_se_plot

Lecture 5: Testing Assumptions for Two-Sample T-Test

For a two-sample t-test, we need to check:

Normality within each group
Equal variances between groups (for standard t-test)
Independent observations

If assumptions are violated:

Welch’s t-test (unequal variances)
Non-parametric alternatives (Mann-Whitney U test)

Practice Exercise 6: Test normality of windward pine needle lengths

qqplots

Note you need to test each groups separately…

# Assuming your dataframe is called df
pine_mean_se_plot

Practice Exercise 7: Test normality of windward pine needle lengths

qqplots

Note you need to test each groups separately…

# how do you make separate dataframes to do this on?
# Separate data by groups
windward_data <- pine_df %>% filter(wind == "wind")
leeward_data <- pine_df %>% filter(wind == "lee")
head(leeward_data)

# A tibble: 6 × 6
  date    group       n_s   wind  tree_no length_mm
  <chr>   <chr>       <chr> <chr>   <dbl>     <dbl>
1 3/20/25 cephalopods n     lee         1        20
2 3/20/25 cephalopods n     lee         1        21
3 3/20/25 cephalopods n     lee         1        23
4 3/20/25 cephalopods n     lee         1        25
5 3/20/25 cephalopods n     lee         1        21
6 3/20/25 cephalopods n     lee         1        16

Practice Exercise 8: Test normality of windward pine needle lengths

qqplots

Note you need to test each groups separately…

# QQ Plot for windward group
qqPlot(windward_data$length_mm, 
       main = "QQ Plot for Windward Pine Needles",
       ylab = "Sample Quantiles")

[1] 21 22

Practice Exercise 9: Test normality of windward pine needle lengths

Shapiro-Wilk test

Note you need to test each groups separately…

# Shapiro-Wilk test for windward group
shapiro_windward <- shapiro.test(windward_data$length_mm)
print("Shapiro-Wilk test for windward data:")

[1] "Shapiro-Wilk test for windward data:"

print(shapiro_windward)


    Shapiro-Wilk normality test

data:  windward_data$length_mm
W = 0.96062, p-value = 0.451

Practice Exercise 10: Test normality of windward pine needle lengths

qqplots

Note you need to test each groups separately…

# You can also test the leeward group
# QQ Plot for leeward group
qqPlot(leeward_data$length_mm, 
       main = "QQ Plot for Leeward Pine Needles",
       ylab = "Sample Quantiles")

[1]  4 16

Practice Exercise 11: Test normality of windward pine needle lengths

Shapiro-Wilk test

Note you need to test each groups separately…

# Shapiro-Wilk test for leeward group
shapiro_lee <- shapiro.test(leeward_data$length_mm)
print("Shapiro-Wilk test for leeward data:")

[1] "Shapiro-Wilk test for leeward data:"

print(shapiro_lee)


    Shapiro-Wilk normality test

data:  leeward_data$length_mm
W = 0.95477, p-value = 0.3425

Practice Exercise 12: Test Normality at one time

There are always a lot of ways to do this in R

# there are always two ways
# Test for normality using Shapiro-Wilk test for each wind group
# All in one pipeline using tidyverse approach
normality_results <- pine_df %>%
  group_by(wind) %>%
  summarize(
    shapiro_stat = shapiro.test(length_mm)$statistic,
    shapiro_p_value = shapiro.test(length_mm)$p.value,
    normal_distribution = if_else(shapiro_p_value > 0.05, "Normal", "Non-normal")
  )

# Print the results
print(normality_results)

# A tibble: 2 × 4
  wind  shapiro_stat shapiro_p_value normal_distribution
  <chr>        <dbl>           <dbl> <chr>              
1 lee          0.955           0.343 Normal             
2 wind         0.961           0.451 Normal

Practice Exercise 13: Test equal variances

Levenes test can be done on the original dataframe

# Method 1: Using car package's leveneTest
# This is often preferred as it's more robust to departures from normality
levene_result <- leveneTest(length_mm ~ wind, data = pine_df)
print("Levene's Test for Homogeneity of Variance:")

[1] "Levene's Test for Homogeneity of Variance:"

print(levene_result)

Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group  1  1.2004 0.2789
      46

Lecture 5: Conducting the Two-Sample T-Test

Now we can compare the mean pine needle lengths between windward and leeward sides.

Ho: μ₁ = μ₂ (The mean needle lengths are equal)

Ha: μ₁ ≠ μ₂ (The mean needle lengths are different)

Deciding between:

Standard t-test (equal variances)
Welch’s t-test (unequal variances)

Note the Levenes Test should be NOT SIGNIFICANT - What is the null hypothesis

levene_result

Levene's Test for Homogeneity of Variance (center = median)
      Df F value Pr(>F)
group  1  1.2004 0.2789
      46

Lecture 5: Conducting the Two-Sample T-Test

Now we can do a two sample TTEST

Calculate t-statistic manually (optional)

YOUR CODE HERE:

t = (mean1 - mean2) / sqrt((s1^2/n1) + (s2^2/n2))

# YOUR TASK: Conduct a two-sample t-test
# Use var.equal=TRUE for standard t-test or var.equal=FALSE for Welch's t-test

# Standard t-test (if variances are equal)
t_test_result <- t.test(length_mm ~ wind, data = pine_df, var.equal = TRUE)
print("Standard two-sample t-test:")

[1] "Standard two-sample t-test:"

print(t_test_result)


    Two Sample t-test

data:  length_mm by wind
t = 8.6792, df = 46, p-value = 3.01e-11
alternative hypothesis: true difference in means between group lee and group wind is not equal to 0
95 percent confidence interval:
 4.224437 6.775563
sample estimates:
 mean in group lee mean in group wind 
          20.41667           14.91667

# Welch's t-test (if variances are unequal)
# YOUR CODE HERE

Lecture 5: Interpreting and Reporting Two-Sample T-Test Results

Interpret the results of the two-sample t-test

What can we conclude about the needle lengths on windward vs. leeward sides?

How to report this result in a scientific paper:

“A two-tailed, two-sample t-test at α=0.05 showed [a significant/no significant] difference in needle length between windward (M = …, SD = …) and leeward (M = …, SD = …) sides of pine trees, t(…) = …, p = ….”

Lecture 5: Interpreting and Reporting Two-Sample T-Test Results

Interpret the results of the two-sample t-test

What can we conclude about the needle lengths on windward vs. leeward sides?

How to report this result in a scientific paper:

Final Activity: Assumptions of Parametric Tests

Common assumptions for t-tests:

Normality: Data comes from normally distributed populations
Equal variances (for two-sample tests)
Independence: Observations are independent
No outliers: Extreme values can influence results

What can we do if our data violates these assumptions?

Alternatives when assumptions are violated:

Data transformation (log, square root, etc.)
Non-parametric tests

Summary and Conclusions

In this activity, we’ve:

Formulated hypotheses about pine needle length
Tested assumptions for parametric tests
Conducted one-sample and two-sample t-tests
Visualized data using appropriate methods
Learned how to interpret and report t-test results

Key takeaways:

Always check assumptions before conducting tests
Visualize your data to understand patterns
Report results comprehensively
Consider alternatives when assumptions are violated